Video Blackjack Odds

Hello. I play a video blackjack game. There is one deck per round - the dealer gets dealt one hand and the player gets dealt 7 hands. There is a bonus for the player to be dealt four blackjacks in one game.

I have been dealt 3 blackjacks out of the 7 hands but never 4. This seems very unlikely. Can anyone give me the odds of being dealt four blackjacks among 7 hands using a single deck? Thank you!

-Sarah

Quote: littlelostbunny

Hello. I play a video blackjack game. There is one deck per round - the dealer gets dealt one hand and the player gets dealt 7 hands. There is a bonus for the player to be dealt four blackjacks in one game.

I have been dealt 3 blackjacks out of the 7 hands but never 4. This seems very unlikely. Can anyone give me the odds of being dealt four blackjacks among 7 hands using a single deck? Thank you!

-Sarah

maybe it's a tease like supposedly happens in slots

people are going to want to know more details. This could be class II gaming for all we know. If class III, make sure you are not being paid '2 for 1' [which is 1:1] for a regular blackjack. That lousy payoff is very common for video BJ and is another thing that would make any other questions mute

PS: welcome Sarah!

Quote: littlelostbunny

Hello. I play a video blackjack game. There is one deck per round - the dealer gets dealt one hand and the player gets dealt 7 hands. There is a bonus for the player to be dealt four blackjacks in one game.

I have been dealt 3 blackjacks out of the 7 hands but never 4. This seems very unlikely. Can anyone give me the odds of being dealt four blackjacks among 7 hands using a single deck? Thank you!

-Sarah

Hey Sarah, and welcome to the forums!

The odds of getting 1 blackjack in a single deck are: P(BJ) = P(A)*P(10) = (4/52)*(16/51) = ~24%

The odds of getting a 2nd one, given the first one and assuming every hand is playing from the same deck (sometimes on video blackjack each hand is from it's own deck) would be: P(BJ2) = P(A)*P(10) = (3/50)*(15/49) = ~18%

Thus,

P(BJ3) = (2/48)*(14/47) = ~12%
P(BJ4) = (1/46)*(13/45) = ~0.6%

So given that you got the first 3, the last 4 hands would each have about a half a percent chance of getting the 4th.

P(4 BJ's) = P(BJ)*(PBJ2)*P(BJ3)*P(BJ4) = .24*.18*.12*.006 = .0000311, or about 1 in 32,000.

I did this quickly so there's a small chance for error but I think that should at least be "close?"

Quote: Romes

Hey Sarah, and welcome to the forums!

The odds of getting 1 blackjack in a single deck are: P(BJ) = P(A)*P(10) = (4/52)*(16/51) = ~24%

The odds of getting a 2nd one, given the first one and assuming every hand is playing from the same deck (sometimes on video blackjack each hand is from it's own deck) would be: P(BJ2) = P(A)*P(10) = (3/50)*(15/49) = ~18%

Thus,

P(BJ3) = (2/48)*(14/47) = ~12%
P(BJ4) = (1/46)*(13/45) = ~0.6%

So given that you got the first 3, the last 4 hands would each have about a half a percent chance of getting the 4th.

P(4 BJ's) = P(BJ)*(PBJ2)*P(BJ3)*P(BJ4) = .24*.18*.12*.006 = .0000311, or about 1 in 32,000.

I did this quickly so there's a small chance for error but I think that should at least be "close?"

The odds of getting bj is about 4.8%, your math was off by a factor of 10 and you forgot to multiply by 2, cause there are 2 ways for every A,T to make a Bj.

4 out of 7 hands being bj is a hypergeometric problem, can probably search for a online calculator and figure it out.

Couldn't you do something like determine the odds of getting 4 (consecutive) blackjacks, of course including EOR, then multiply by 7/4 (or something like that)?

ie:

[(4/52 * 14/51 * 2) x (3/50 * 13/49 * 2) x (2/48 * 12/47 * 2) x (1/46 * 11/45 * 2)] * 7/4 = 5.32066049e-7

1/(5.32066049e-7) = 1879465 ~ 1 in 1,879,465 rounds you'll get 4 blackjacks.

???


Idk if that's right or not. But maybe.